Counting 1..2..3!

At the beginning of today's class, the class was divided into 5 groups of 4. Each group were given the same 5 questions on counting. As seen below. I will now explain briefly how to get these answers and why.Solutions and Explanations

1.a) As seen in question one the answer is 21 terms. Determining how many terms there are is defined by (n +1) , n is the exponent after the binomial expansion. So ( 9+1) =10 and (10 +1)=11. Therefore 10+11 = 21 terms.

1.b) The answer is (x +y) ^10. This is because in order for you to have a middle term, you need to have the same amount of numbers before and after that middle number, and this can only be done with an odd number of terms, but since our formula is (n+1), in finding how many terms, in a binomial expansion it would have to be an even number. (x +y) ^10, has 11 terms therefore having 5 before and 5 after. The middle term would be 6 in this expansion.

c) As said above, If "n" is an even number.

2. The answer is there is no middle term because the "n" was an odd number.

3. a) self explanatory. (18)(10)(20)(10)(20)(10) <given to us,and 10 stands for numbers you can choose between 0-9.

b) KEY WORD IS NEW. So without restrictions it would look like this (26)(10)(26)(10)(26)10) but since how many new ones are formed you would subtract the answer from, part a, from the "no restrictions" to get all the new codes.

4. It is 7P4. It is pick, because order matters, enemies do not want to sit beside enemies, or window sitters. 7 stands for the empty seats the can pick from, and 4 stands for the people who will be sitting in their picked spots.

5. a) 18C6 because order does not matter. 18 stands for number of people. 6 stands for people choosen to be on committee.

b) 8C4(10C2), exactly 4 boys, so since there are 8 boys they have to choose exactly four, and remaining 2 people will have to be chosen from the 10 girls, to make the committee of 6 people.

c) 10C4(8C2)Same explanation as above, but exactly 4 girls have to be chosen from the 10 girls , then the other 2 people will have to be chosen from the 8 boys, to form the committee.

scribe post: counting

Hello guys... This is Richard and it's my third scribe post. But before we proceed to our lesson for today, let's take a look to the notes on last thursday (November 16) since there's no scribe that day.

COUNTING (a.k.a. COMBINATORICS)

The Fundamental Principle of Counting

If there are m ways to do the first thing and n ways to do a second thing then there are m • n ways to do both things.

example:

How many different outfits can be made if a person has 3 pairs of pants and 4 shirts?

solution:

3 ways to choose pants

4 ways to choose shirts


۞ 3 x 4 = 12 ways to choose both

Factorial Notation

Definition:

n! = n (n-1) (n-2) (n-3) . . . 3, 2, 1

0! = 1

examples:

4! = 4 • 3 • 2 • 1

= 24

6! = 6 • 5 • 4 • 3 • 2 • 1

= 720

Note: FACTORIAL NOTATION IS UNDEFINED FOR NEGATIVE NUMBERS

n! = n (n-1)!

Permutation

An order arrangement of objects.

FORMULA

nPr = n!/ (n-r)!

n is the number of objects to pick from

r is the number of object to be arrange

nPr is read as " n Pick r"

nPr given a set of n objects. How many orderd arrangements can be made using only r of them at a time.

examples:

There are 8 horses in a race. In how many different ways can they finish 1st, 2nd, 3rd.

solution:

(by fundamental principle of counting)

8 • 7 • 6

- - - = 336

1 2 3

(by Permutation)

8P3 = 8! / (8-3)!

= 8!/ 5!

= 336

Permutation of Non-distinguishable objects

The number of ways to arrange "n" objects that contain K1, K2, K3. . . .etc. Set of non-distinguishable objects is given by:

n!

__________

K1! K2! K3!...

example:

How many different words can be made from the letters in the words:

(a) book (b) Mississippi

(a) # letters = 4

# of O's = 2

= 4! / 2!

= 12

(b) # letters = 11

# of I's = 4

# of S's = 4

# of P's = 2

= 11! / (2! 4! 4!)

November 21, 2006

In the beginning of the class, Mr.K give us the following problems without using a calculator.

Simplify

(a) 8! / 4!

8•7•6•5•4!

_______ 4! is reduce

4!

so we left with 8 • 7 • 6 • 5

= 1680

(b) 7! - 5!

7 • 6 • 5! - 5!

5! (7 • 6 - 1) factoring by 5!

120 (41) = 4920

(c) 10! 4! / 8! 6!

10•9•8! 4!

_______ 8! & 4! are reduce

8! 6•5•4!

10•9

____ = 3

6•5

Write using factorial

(a) 5•4•3•2•1

= 5!

(b) 5•4•3

= 5! / 2!

(c) 20•19•18•17

=20! / 16!

Simplify

(a) (n+1)! / n!

(n+1) n!

______ from the definition of factorial notation n! = n ( n-1)!

n!

= n + 1

(b) (n-1)! / n!

(n-1)!

_____

n (n-1)!

= 1 / n

(c) (n+1) (n-1) / (n!)²

(n+1) n! (n-1)!

__________

n! n (n-1)!

= (n+1) / n

(d)

(2n-1)! (n+1)!

__________

(2n+1)! (n-1)!

(2n-1)! (n+1)( n ) (n-1)!

_________________

(2n+)(2n)(2n-1)! (n-1)!

n(n+1)

_______ n is reduce

2n(2n+1)

n+1

____

4n+2

SOLVE:

(a) 3! (n-1)!

_______ = 72

(n-3)!

(n-1) ! 72

____ = __

(n-3)! 6

(n-1)(n-2) = 12 (n-3)! is reduce.

n² -3n +2 = 12

n²-3n-10 = 0

(n-5) (n+2) = 0

n= 5 n= -2

-2 is rejected because the factorial of negative numbers is undefined.

Always indicate if it is accepted or reject.

(b) (2n-1)!

________ = 10

2! (2n -3)

(2n-1) (2n-2) (2n-3)

______________ = 20

(2n-3)

(2n-1) (2n-2) = 20

4n² -6n - 18 = 0

2n² -3n - 9 = 0 it is divided by 2

(2n+3) (n-3)

n= -3/2 n = 3

-3/2 is rejected

I'm sorry guys if this is not nice..I'll fix it later. I just can't upload the images.

By the way the next scribe will be John.

Using the Counting Principle with Restrictions

Here is the assignment we began in class today ... finish it as well as Exercise #32 for class tomorrow. The answers are provided below. WARNING: Number 11 is difficult.



Answers:
1. 120

2. 20 160

3. (a) 840
(b) 240
(c) 144

4. (a) 2160
(b) 1260
(c) 660

5. 144

6. 4320

7. (a) 4536
(b) 2240
(c) 952

8. (a) 5040
(b) 720

9. (a) 576
(b) 72

10. (a) 891
(b) 445
(c) 179

11. 240

12. (a) 10 416
(b) 5082